Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Centre of Mass, Linear momentum, Collision
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision
A block of mass m is positioned on a triangular block of mass M ,...
A block of mass is positioned on a triangular block of mass , which in turn rests on a horizontal surface as shown in the figure. Assuming all surfaces are frictionless, determine the velocity of the triangular block when the smaller block reaches the bottom end.
Options
- A\left[\dfrac{2M^2 gh\cos^2\alpha}{(M + m)(m + M\sin^2\alpha)}\right]^{1/2}
- B\left[\dfrac{2m^2 gh\sin^2\alpha}{(M + m)(M + m\cos^2\alpha)}\right]^{1/2}
- C\left[\dfrac{2m^2 gh\cos^2\alpha}{(M + m)(M - m\sin^2\alpha)}\right]^{1/2}
- D\left[\dfrac{2m^2 gh\cos^2\alpha}{(M + m)(M + m\sin^2\alpha)}\right]^{1/2}
Correct answer
D. \left[\dfrac{2m^2 gh\cos^2\alpha}{(M + m)(M + m\sin^2\alpha)}\right]^{1/2}
Step-by-step solution
Let be the velocity of the triangular block of mass towards the left.
Let be the velocity of the block of mass relative to the triangular block, directed down the incline.
The velocity of the block relative to the ground is .
Taking the right direction as positive y\vec{V} = -V \hat{i}\vec{v}_r = v_r \cos\alpha \hat{i} - v_r \sin\alpha \hat{j}\vec{v} = (v_r \cos\alpha - V) \hat{i} - v_r \sin\alpha \hat{j}M(-V) + m(v_r \cos\alpha - V) = 0m v_r \cos\alpha = (M + m) Vv_r = \dfrac{(M + m) V}{m \cos\alpha}mgh = \dfrac{1}{2} M V^2 + \dfrac{1}{2} m v^2v^2 = (v_r \cos\alpha - V)^2 + (-v_r \sin\alpha)^2 = v_r^2 + V^2 - 2 v_r V \cos\alphav_rv^2v^2 = \left( \dfrac{(M + m) V}{m \cos\alpha} \right)^2 + V^2 - 2 \left( \dfrac{(M + m) V}{m \cos\alpha} \right) V \cos\alphav^2 = V^2 \left[ \dfrac{(M + m)^2}{m^2 \cos^2\alpha} +\dfrac{2(M + m)}{m} \right]
Substituting into the energy equation:
\dfrac{1}{2}\dfrac{1}{2}
Multiplying by and expanding:
2mgh = M V^2 + m V^2 \left[ \dfrac{(M + m)^2}{m^2 \cos^2\alpha} +1 - \dfrac{2(M + m)}{m} \right]<br>2mgh = V² \left[ M + \dfrac{(M + m)^2}{m \cos^2\alpha} + m - 2(M + m) \right]2mgh = V^2 \left[ (\dfrac{(M + m)^2}{m \cos^2\alpha} \right]
2mgh = V^2 (M + m) \left[\dfrac{M + m}{m \cos^2\alpha} -1 \right]
2mgh = V² (M + m) \left[ \dfrac{M + m \sin^2\alpha}{m \cos^2\alpha} \right]VV^2 = \dfrac{2m^2 gh \cos^2\alpha}{(M + m)(M + m \sin^2\alpha)}V = \left[ \dfrac{2m^2 gh \cos^2\alpha}{(M + m)(M + m \sin^2\alpha)} \right]^{1/2}
Let be the velocity of the block of mass relative to the triangular block, directed down the incline.
The velocity of the block relative to the ground is .
Taking the right direction as positive y\vec{V} = -V \hat{i}\vec{v}_r = v_r \cos\alpha \hat{i} - v_r \sin\alpha \hat{j}\vec{v} = (v_r \cos\alpha - V) \hat{i} - v_r \sin\alpha \hat{j}M(-V) + m(v_r \cos\alpha - V) = 0m v_r \cos\alpha = (M + m) Vv_r = \dfrac{(M + m) V}{m \cos\alpha}mgh = \dfrac{1}{2} M V^2 + \dfrac{1}{2} m v^2v^2 = (v_r \cos\alpha - V)^2 + (-v_r \sin\alpha)^2 = v_r^2 + V^2 - 2 v_r V \cos\alphav_rv^2v^2 = \left( \dfrac{(M + m) V}{m \cos\alpha} \right)^2 + V^2 - 2 \left( \dfrac{(M + m) V}{m \cos\alpha} \right) V \cos\alphav^2 = V^2 \left[ \dfrac{(M + m)^2}{m^2 \cos^2\alpha} +\dfrac{2(M + m)}{m} \right]
Substituting into the energy equation:
\dfrac{1}{2}\dfrac{1}{2}
Multiplying by and expanding:
2mgh = M V^2 + m V^2 \left[ \dfrac{(M + m)^2}{m^2 \cos^2\alpha} +1 - \dfrac{2(M + m)}{m} \right]<br>2mgh = V² \left[ M + \dfrac{(M + m)^2}{m \cos^2\alpha} + m - 2(M + m) \right]2mgh = V^2 \left[ (\dfrac{(M + m)^2}{m \cos^2\alpha} \right]
2mgh = V^2 (M + m) \left[\dfrac{M + m}{m \cos^2\alpha} -1 \right]
2mgh = V² (M + m) \left[ \dfrac{M + m \sin^2\alpha}{m \cos^2\alpha} \right]VV^2 = \dfrac{2m^2 gh \cos^2\alpha}{(M + m)(M + m \sin^2\alpha)}V = \left[ \dfrac{2m^2 gh \cos^2\alpha}{(M + m)(M + m \sin^2\alpha)} \right]^{1/2}